Answer:
E=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
[tex]E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}[/tex]
Substitute x=a and R=a
Then, we get
[tex]E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}[/tex]
[tex]E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}[/tex]
[tex]E=\frac{KQa}{2\sqrt 2a^3}[/tex]
[tex]E=\frac{KQ}{2\sqrt 2a^2}[/tex]
Where K=[tex]9\times 10^9 Nm^2/C^2[/tex]
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]
The magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]
The electric field due to a charged ring E is given by
E = Qz/4πε₀[√(z² + R²)]³ where
Given that for this ring R = a and z = a, substituting these values into E, the magnitude of the electric field at a is given by
E = Qz/4πε₀[√(z² + R²)]³
E = Qa/4πε₀[√(a² + a²)]³
E = Qa/4πε₀[√(2a²)]³
E = Qa/4πε₀[2√2a³]
E = Q/[8πε₀√2a²]
E = Q/[8√2πε₀a²]
So, the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]
Learn more about magnitude of the electric field due to a charged ring
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